__Question 4.16__:

A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball ?

__Solution__:

Maximum horizontal distance,

*R*= 100 m

The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45°, i.e.,

*θ*= 45°.

The horizontal range for a projection velocity

*v*, is given by the relation:

R = u

^{2}Sin 2θ / g

100 = u

^{2}Sin 90

^{0}/ g

u

^{2}/ g = 100 ....(i)

The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity

*v*is zero at the maximum height

*H*.

Acceleration,

*a = –g*

Using the third equation of motion:

v

^{2}- u

^{2}= -2gH

H = u

^{2}/ 2g = 100 / 2 = 50 m

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